Problem: $f(x) = -2x + 3$ for all real numbers. What is $f^{-1}(x)$, the inverse of $f(x)$ ? 2 4 6 8 \llap{-}4 \llap{-}6 \llap{-}8 2 4 6 8 \llap{-}4 \llap{-}6 \llap{-}8 $f(x)$ $f^{-1}(x)$
Explanation: $y = f(x)$ , so solving for $x$ in terms of $y$ gives $x=f^{-1}(y)$ $f(x) = y = -2x+3$ $y-3 = -2x$ $-\dfrac{y}{2}+\frac{3}{2} = x$ $x = -\dfrac{y}{2}+\frac{3}{2}$ So we know: $f^{-1}(y) = -\dfrac{y}{2}+\frac{3}{2}$ Rename $y$ to $x$ $f^{-1}(x) = -\dfrac{x}{2}+\frac{3}{2}$ Notice that $f^{-1}(x)$ is just $f(x)$ reflected across the line $y=x$.